Module 2 : Electrostatics
Lecture 7 : Electric Flux
Objectives
In this lecture you will learn the following
Concept of flux and calculation of eletric flux throught simple geometrical objects
Gauss's Law of electrostatics
Applications of Gauss's Law to Calculate electric field due to a few symmetric charge distributions.
Electric Flux
The concept of flux is borrowed from flow of water through a surface. The amount of water flowing through a
surface depends on the velocity of water, the area of the surface and the orientation of the surface with respect
to the direction of velocity of water.
Though an area is generally considered as a scalar, an element of area may be considered to be a vector
because
It has magnitude (measured in m
).
If the area is infinitisimally small, it can be considered to be in a plane. We can then associate a
direction with it.
For instance, if the area element lies in the x-y plane, it can be considered to be directed along the z-direction.
(Conventionally, the direction of the area is taken to be along the outward normal.)
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is chosen to represent the area in some convenient unit and its
In the figure above, the length of the vector
direction is taken to be along the outward normal to the area.
We define the flux of the electric field through an area
If
to be given by the scalar product
is the angle between the electric field and the area vector
For an arbitrary surface S, the flux is obtainted by integrating over all the surface elements
If the electric field is uniform, the angle
is constant and we have
Thus the flux is equal to the product of magnitude of the electric field and the projection of area perpendicular
to the field.
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Unit of flux is N-m
it.
/C. Flux is positive if the field lines come out of the surface and is negative if they go into
Solid Angle :
The concept of solid angle is a natural extension of a plane angle to three dimensions. Consider an area element
dS at a distance
from a point P. Let be the unit vector along the outward normal to .
The element of the solid angle subtended by the area element at P is defined as
where is the projection of along a direction perpendicular to . If is the angle between
and , then,
Solid angle is dimensionless. However, for practical reasons it is measured in terms of a unit called steradian
(much like the way a planar angle is measured in terms of degrees).
The maximum possible value of solid angle is , which is the angle subtended by an area which encloses the
point P completely.
Example
A right circular cone has a semi-vertical angle
. Calculate the solid angle at the apex P of the cone.
Solution :
The cap on the cone is a part of a sphere of radius R, the slant length of the cone. Using spherical polar
coordinates, an area element on the cap is
sin
, where
is the polar angle and
is the
azimuthal angle. Here,
goes from 0 to
while
goes from 0 to
.
Thus the area of the cap is
Thus the solid angle at P is
Exercise
Calculate the solid angle subtended by an octant of a sphere at the centre of the sphere.
(Ans.
)
The flux per unit solid angle is known as the intensity .
Example 3
An wedge in the shape of a rectangular box is kept on a horizontal floor. The two triangular faces and the
rectangular face ABFE are in the vertical plane. The electric field is horizontal, has a magnitude
N/C
and enters the wedge through the face ABFE, as shown. Calculate the flux through each of the faces and
through the entire surface of the wedge.
The outward normals to the triangular faces AED, BFC, as well as the normal to the base are perpendicular to
. Hence the flux through each of these faces is zero. The vertical rectangular face ABFE has an area 0.06 m
. The outward normal to this face is perpendicular to the electric field. The flux is entering through this face
and is negative. Thus flux through ABFE is
To find the flux through the slanted face, we need the angle that the normal to this face makes with the
horizontal electric field. Since the electric field is perpendicular to the side ABFE, this angle is equal to the angle
. The area of the slanted face ABCD is 0.1 m
between AE and AD, which is
. Thus the flux
through ABCD is
The flux through the entire surface of the wedge is
Example 4
Calculate the flux through the base of the cone of radius
.
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Solution :
The flux entering is perpendicular to the base. Since the outward normal to the circular base is in the opposite
sense, the flux is negative and is equal to the product of the magnitude of the field and the area of the base,
The flux, therefore is,
.
Example 5
Calculate the flux coming out through the curved surface of the cone in the above example.
Solution :
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Consider a circular strip of radius
through the strip and the vector
at a depth
is
, where
element along the slope, the area of the strip is
We have,
to
is the semi-angle of the cone. If
is the length
. Thus,
. Further, r = h tan
, so that
Integrating from
from the apex of the cone. The angle between the electric field
Substituting, we get
, the height of the cone, the outward flux is
.
Example 6
A charge
is located at the center of a sphere of radius
. Calculate the flux going out through the surface
of the sphere.
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By Coulomb's law, the field due to the charge
is radial and is given on the surface of the sphere by,
The direction of the area vector
, is also radial at each point of the surface
The integral over
is equal to the surface area of the sphere, which is,
surface of the sphere is
. The flux
. Thus the flux out of the
GAUSS'S LAW - Integral form
The flux calculation done in Example 4 above is a general result for flux out of any closed surface, known as
Gauss's law.
Total outward electric flux
through a closed surface
is equal to
times the charge enclosed
by the volume defined by the surface
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Mathematicaly, the surface integral of the electric field over any closed surface is equal to the net charge
enclosed divided by
The law is valid for arbitry shaped surface, real or imaginary.
Its physical content is the same as that of Coulomb's law.
In practice, it allows evaluation of electric field in many practical situations by forming imagined
surfaceswhich exploit symmetry of the problem. Such surfaces are called Gaussian surfaces .
GAUSS'S LAW - Differential form
The integral form of Gauss's law can be converted to a differential form by using the divergence theorem. If
is the volume enclosed by the surface S,
If
is the volume charge density,
Thus we have
Applications of Gauss's Law
Field due to a uniformly charged sphere of radius with a charge
Gaussian surface is a cylinder of radius
and length
.
By symmetry, the field is radial. Gaussian surface is a concentric sphere of radius
the Gaussian surface is parallel to the field at every point. Hence For ,
. The outward normals to
so that
The field outside the sphere is what it would be if all the charge is concentrated at the origin of the sphere.
For , a fraction of the total charge is enclosed within the gaussian surface, so that
The field inside is
Exercise
Find the electric field both inside and outside a spherical shell of radius carrying a uniform charge .
Example
Find the electric field inside a sphere of radius which carries a charge density where
origin and is a constant.
is the distance from the
Solution :
By symmetry the field is radial. Take the gaussian surface to be a sphere of radius . The flus is . The charge
enclosed by the gaussian surface is
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Thus
where
is a unit vector perpendicular to the line,directed outward for positive line charge and inward for
negative line charge.
Field due to an infinite charged sheet with surface charge density
Choose a cylindrical Gaussian pillbox of height
radius
(with
above the sheet and
below the sheet) and
.
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The amount of charge enclosed is area times the surface charge density, i.e.,
. By symmetry, the
field is directed perpendicular to the sheet, upward at points above the sheet and downward for points below.
There is no contribution to the flux from the curved surface. The flux from the two end faces is
each,
i.e. a total outward flux of
. Hence
so that
where
is a unit vector perpendicular to the sheet, directed upward for points above and downwards for points
below (opposite, if the charge density is negative).
Field due to a uniformly charged sphere of radius
with a charge
By symmetry, the field is radial. Gaussian surface is a concentric sphere of radius
the Gaussian surface is parallel to the field
. The outward normals to
at every point. Hence
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For
,
so that
The field outside the sphere is what it would be if all the charge is concentrated at the origin of the sphere.
For
, a fraction
The field inside is
of the total charge is enclosed within the gaussian surface, so that
Exercise 1
Find the electric field in the region between two infinite parallel planes carrying charge densities
and
.
Exercise 2
Find the electric field both inside and outside a spherical shell of radius
carrying a uniform charge
.
Exercise 3
Find the electric field both inside and outside a long cylinder of radius
density
carrying a uniform volume charge
.
(Hint : Take the gaussian surface to be a finite concentric cylinder of radius
(with
and
), as
shown)
Example 7
Find the electric field inside a sphere of radius
distance from the origin and
which carries a charge density
where
is a constant.
Solution :
By symmetry the field is radial. Take the gaussian surface to be a sphere of radius
. The flus is
is the
.
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The charge enclosed by the gaussian surface is
Thus
(what is the dimension of
?)
Exercise 4
A very long cylinder carries a charge density
Find the electric field at a distance
.
, where
(Ans.
is the distance from the axis of the cylinder.
)
Exercise 5
A charge
(Ans.
is located at the center of a cube of side
. Find the flux through any of the sides.
)
Example 8
Find the flux through the curved surface of a right circular cone of base radius
. The cone has no charge and the electric field is normal to the base.
in an external electric field
Solution :
In Example 4, we calculated the flux through the base to be
. As the cone does not contain any
charge, by Gauss's law, the flux through the curved surface must be
,
which is the result we obtained in Example 4.
Recap
In this lecture you have learnt the following
Electric flux through a surface is surface integral of normalcomponent of electric field through the
surface.
Gauuss's of electrostatics states that the flux of electric field from any closed surface is proportional to
the charge enclosed by the surface. Both the integral and differential form of Gauss's law were studied.
Gauss's law helps us to deternmine electric field due to charge distributions having spatial symmetry.
Fields due to distributions showing spherical and cylidrical symmetry were studied.
Electric field due a charged sheet was obtained using Gauss's law.